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For a function $f(t)$ Laplace transform is defined as $F(s)=\int_0^{\infty} f(t)e^{-st}dt$.

I have to show the property that the Laplace transform of $f(t)\over t$ is $\int _s^\infty F(s')ds'$.

I tried the substitution $k={1\over t}$ but then I end up with $\int_0^\infty {e^{-sk} f(1/k)\over k} dk$.

In $\int _s^\infty F(s')ds'$ is $s'$ a dummy variable or is it the derivative of $s$ in$ F(s)$.
Can someone please help me to show this

sam_rox
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2 Answers2

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$s'$ is a dummy variable, not the derivative of $s$ with respect to anything. In any case, it can sometimes be helpful to start at the far side of the equals sign. We have

\begin{align} \int_s^{\infty}F(s')ds' & = \int_s^{\infty}\int_0^{\infty}e^{-s' t}f(t)dtds'\\ & = \int_0^{\infty}\int_s^{\infty}e^{-s't}f(t)ds'dt\\ & = \int_0^{\infty}\left.-\frac{1}{t}e^{-s't}f(t)\right|_s^{\infty}dt\\ & = \int_0^{\infty}e^{-st}\frac{f(t)}{t}dt\\ & = \mathcal{L}\left[\frac{f(t)}{t}\right] \end{align}

solitaireartist
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Hint: compute both $$ \frac d{ds} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt \\ \lim _{s\to\infty} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt $$

mookid
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  • $\frac d{ds} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt=-F(s)$.I don't understand what I should do with the limit – sam_rox Jun 19 '14 at 08:43
  • you conclude that $\int_0^{\infty} f(t)/t e^ {-st} dt = -\int f(s')ds' + C$. Then, you need to compute $C$. – mookid Jun 19 '14 at 08:47
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    I haven't really understood this.As I understand $L({f(t)\over t})=\int_0^\infty \frac{f(t)}{t}e^ {-st} dt$.Differentiating both sides of this with respect to s, $d/ds(L({f(t)\over t}))=-F(s)$. Then integrating $L({f(t)\over t})=-\int F(s) ds +C$.Is this right so far?How can I go from here – sam_rox Jun 19 '14 at 08:57
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    You need to compute C. The limit when $s\to\infty$ of LHS and RHS is easy to compute. – mookid Jun 19 '14 at 09:04
  • Is $\int F(s)ds$ and $\int f(s')ds'$ the same.Why do I have to compute a limit – sam_rox Jun 19 '14 at 09:29